Problem: Complete the square to solve for $x$. $x^{2}+6x-7 = 0$
Solution: Begin by moving the constant term to the right side of the equation. $x^2 + 6x = 7$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $6$ , half of it would be $3$ , and squaring it gives us ${9}$ $x^2 + 6x { + 9} = 7 { + 9}$ We can now rewrite the left side of the equation as a squared term. $( x + 3 )^2 = 16$ Take the square root of both sides. $x + 3 = \pm4$ Isolate $x$ to find the solution(s). $x = -3\pm4$ So the solutions are: $x = 1 \text{ or } x = -7$ We already found the completed square: $( x + 3 )^2 = 16$